EasySpin forum

Hi All,

I don't have a lot of experience on a exchange coupled system or more complicated spin system, so I might have a lot of basic questions on that.
I have questions for building up hyperfine matrix for an exchange coupled system since I didn't see a lot of discussion and case study here.
I have made a stable biradical compound that likely possess strong exchange coupling at ground state with isotropic EPR spectrum collected. I am having trouble simulating it for building up proper hyperfine matrix.
It is a hetero biradical, one's hyperfine comes from a proton, the other unpaired spin is shared by 4 nitrogens which can be grouped into 2 sets of equivalent nitrogen (verdazyl radical). Each set of N gives one hyperfine constant. So this is a 2 electron, 5 nuclei system.
Below is the system I built with exchange coupling without A matrix yet:
Sys.g=[2.0039,2.0041]
Sys.lw=0.15
Sys.S=[0.5 0.5]
Sys.Nucs='N,N,N,N'
Sys.J=-7790000

I am starting with a simple case that I only focus on the dominant hyperfine come from 4 N. The system is then a 2 electron, 4 nuclei system. What confused me is the discussion of spin system in the manual.
Let's say the two sets of N give hyperfine constant 15 and 20, respectively. According to the manual, for a 2 electron 2 nuclei system, the hyperfine matrix is like [10 10 -20 30 40 50; 1 1 -2 3 4 5].
So my first question is, for each block, does 3 of the number stands for hyperfine for each spin? For example, for '10 10 -20', is that the hyperfine for the first electron on first nucleus?
The second question, if it's right, then for an isotropic spectrum, the 3 number should be identical?
The third question is about my own system. With 4 Nitrogen in two sets, and 2 electrons, below is the building for an isotropic A matrix if I ignore the hyperfine interaction between nuclei and another spin:
[15 15 15 0 0 0;15 15 15 0 0 0; 20 20 20 0 0 0; 20 20 20 0 0 0]
Is that right?
If I do not ignore that, my matrix will be like:
[15 15 15 x x x; 15 15 15 x x x; 20 20 20 y y y; 20 20 20 y y y]
Is that right?
Or, since it is isotropic, can I downsize the matrix to:
[15 x;15 x;20 y;20 y] ?

The last question is, if my previous guess is right, is easyspin capable of specifying the nucleus? For example, I want the first 3 number to be those on the first nucleus, and the other 3 is for the other nucleus, since I'm working on a hetero biradical? I'm not sure if the first 3 number and the latter 3 can exchange position without problem?

My full spin system is actually beyond the simple model I built here, I also have two protons on the verdazyl radical part, and one from another radical. If I want to have a very good simulation then it will cost more computing effort, hence I need to minimize any mistake in simulation so I can take a chance understanding this issue.

I appreciate your suggestion on this, thank you!

Best,
Ju

Sys.A is matrix that consists of blocks:

if (ei-nj) is the coupling between electron i and nucelus j, Sys.A looks like

[ [e1-n1] [e2-n1];
[e1-n2] [e2-n2];
[e1-n3] [e2-n3];
[e1-n4] [e2-n4];
[e1-n5] [e2-n5]; ]

each of these blocks can be one, two or three numbers, or a 3x3 matrix. but they all need to be in the same format.
you can NOT interchange the blocks.

e1 refers to the first electron in Sys.S and Sys.g and so on
n1 refers to the first nucleus in the string Sys.Nucs.

If there is no HF coupling between the electrons of center 1 and the nuclei of center 2 and vice versa, then half of each row above has to be 0;

You can easily test your understanding: use two different g values and to different HF couplings and no J-coupling. It should then be easy to predict the obtained spectrum.

Best,
Nino

nwili wrote:Sys.A is matrix that consists of blocks:

if (ei-nj) is the coupling between electron i and nucelus j, Sys.A looks like

[ [e1-n1] [e2-n1];
[e1-n2] [e2-n2];
[e1-n3] [e2-n3];
[e1-n4] [e2-n4];
[e1-n5] [e2-n5]; ]

each of these blocks can be one, two or three numbers, or a 3x3 matrix. but they all need to be in the same format.
you can NOT interchange the blocks.

e1 refers to the first electron in Sys.S and Sys.g and so on
n1 refers to the first nucleus in the string Sys.Nucs.

If there is no HF coupling between the electrons of center 1 and the nuclei of center 2 and vice versa, then half of each row above has to be 0;

You can easily test your understanding: use two different g values and to different HF couplings and no J-coupling. It should then be easy to predict the obtained spectrum.

Best,
Nino

Thanks Nino,

Your reply is just in line with my thoughts.